Exams › JEE Main › Physics

Two particles of masses m1 and m2 undergo a head-on collision while moving in the same direction with initial speeds u1 and u2. The interaction lasts for a time interval dt starting at t = 0. If during the collision the speed of the first particle changes as v(t) = u1 + (t/dt)*(v1 - u1), where v1 is its final speed, find the speed of the second particle as a function of time t during the collision.

1. u2 + (m1/m2)*(t/dt)*(u1 - v1)
2. u2 - (m1/m2)*(t/dt)*(u1 - v1)
3. u2 + (m2/m1)*(t/dt)*(u1 - v1)
4. u2 - (t/dt)*(u1 - v1)

Correct answer: u2 + (m1/m2)*(t/dt)*(u1 - v1)

Solution

Momentum conservation at each instant gives the second particle's speed; whatever momentum the first particle loses is gained by the second, so V(t) increases linearly with t.

Related JEE Main Physics questions

• A satellite moving through a region of free space collects interplanetary dust that is initially at rest. If the dust is gathered at the rate dM/dt = αv, then the satellite’s acceleration is
• A block of mass m lies at rest on a frictionless horizontal floor. A uniform rope of mass m/3 has one end attached to the block, and the free end is pulled horizontally by a force F. What is the tension at the midpoint of the rope?
• A block of mass m is attached to a second block of mass M by a light spring of spring constant k. Both blocks lie on a frictionless horizontal surface. At the beginning, the system is at rest and the spring is at its natural length. A constant force F is then applied to the block of mass M to pull it. Determine the force exerted by the block of mass m.
• A thin ring of mass M and radius R is spinning about its symmetry axis with angular speed ω. Two identical small masses, each of mass m, are then lightly fixed at the two opposite ends of a diameter of the ring. The decrease in kinetic energy is
• A sphere rolls on a surface without slipping. Its radius of gyration about an axis through its centre of mass is K, and its radius is R. What fraction of the total kinetic energy is due to rotation?
• A uniform circular ring has mass M and radius R. If a 90° arc of the ring is cut away, the moment of inertia of the remaining portion about an axis through the centre of the ring and perpendicular to its plane becomes kMR². What is the value of k?

⚔️ Practice JEE Main Physics free + battle 1v1 →